Question

What is the axis of symmetry, vertex, and y-intercept for f(x)=0.5x^2-2x-2

Answer 1

`f(x)=0.5x^2-2x-2`

U

For the axis of symmetry, formula is,

`x=(-b)/(2a)`
`\begin{gathered} \text{where a=0.5 and b=-2} \\ x=(-(-2))/(2(0.5))=(2)/(1)=2 \end{gathered}`

Hence, the axis of symmetry is 2.

To find the vertex, f(x)=0,

`\begin{gathered} 0=0.5x^2-2x-2 \\ \text{Multiply both sides by 2} \\ 0=x^2-4x-4 \\ 0=(x^2-4x)-4 \\ 0=(x-2)^2-(4-2_{} \\ 0=(x-2)^2-2 \\ \text{Vertex are (2, -2)} \end{gathered}`

Hence, the vertex are (2, -2).

`y-intercept\text{ is -2}`

Hence, y-intercept is -2.